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2 years ago

(-0.3) + 0.9 find the sum

Mathematics

2 answers:

lesya692 [45]2 years ago

7 0

The answer is 0.6 how it helped

IceJOKER [234]2 years ago

3 0

Answer: (-0.3) + 0.9= 0.6

Step-by-step explanation:

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Nora and Mindy are going to a movie. Mindy forgot her money and is going to pay Nora back later. The cost of the two tickets is

Agata [3.3K]

Answer:

$9.62

Step-by-step explanation:

15.50 + 3.74 = 19.24

19.24 / 2 = 9.62

so Mindy must pay Nora $9.62

6 0

2 years ago

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A website advertises job openings on its website, but job seekers have to pay to access the list of job openings. The website re

nataly862011 [7]

Answer:

The 90% confidence interval would be given by (57.006;62.994)

Step-by-step explanation:

Previous concepts

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

$\bar X=60$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=10$ represent the population standard deviation

n=30 represent the sample size

Assuming the X follows a normal distribution

$X \sim N(\mu, \sigma=10)$

The sample mean $\bar X$ is distributed on this way:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

The confidence interval on this case is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The next step would be find the value of $\z_{\alpha/2}$ , $\alpha=1-0.90=0.1$ , $\alpha/2 =0.05$ and $z_\alpha/2=1.64$

Using the normal standard table, excel or a calculator we see that:

$z_{\alpha/2}=1.64$

Since we have all the values we can replace:

$60 - 1.64\frac{10}{\sqrt{30}}=57.006$

$60 + 1.64\frac{10}{\sqrt{30}}=62.994$

So on this case the 90% confidence interval would be given by (57.006;62.994)

5 0

2 years ago

If f(x) = −3x + 4 and g(x) = 2, solve for the value of x for which f(x) = g(x) is true.

Komok [63]

Answer

2/3

Explanation

For f(x) = g(x) we should equate the two functions to get the value of x.

f(x) = g(x)

−3x + 4 = 2

-3x + 4 - 4 = 2 - 4

-3x = -2

Dividing by -3 on both sides;

(-3x)/-3 = (-2)/-3

x = 2/3

4 0

2 years ago

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PLEASE HELP ME I'LL FAIL MATH 20 POINTS

TiliK225 [7]

Answers:
1. 5(x-6)
2. k-9
3. 10x + 5
4. the quotient of some number and six
5. the quotient of three times a number and five
6. x^5

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2 years ago

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g An irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the

blagie [28]

Answer:

A 90% confidence interval of the true mean is [$119.86, $123.34].

Step-by-step explanation:

We are given that an irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the mean amount of money spent on textbooks was $121.60.

Also, the standard deviation of the population was $6.36.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean amount of money spent on textbooks = $121.60

$\sigma$ = population standard deviation = $6.36

n = sample of students = 36

$\mu$ = population mean

Here for constructing a 90% confidence interval we have used One-sample z-test statistics as we know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $121.60-1.645 \times {\frac{6.36}{\sqrt{36} } }$ , $121.60+1.645 \times {\frac{6.36}{\sqrt{36} } }$ ]

= [$119.86, $123.34]

Therefore, a 90% confidence interval of the true mean is [$119.86, $123.34].

5 0

2 years ago